I was thinking if the 10 ohm resistor (horizontal position in the image) should be removed, because if we analyse it from Wheatstone bridge's point of view, no current flows through it. I maybe wrong. Help appreciated.

Staff: Mentor

Hi sagarbhathwar.

In future, please make use of the Homework Template provided when you begin a new thread in the Homework sections.

I wouldn't recommend removing that horizontal resistor as what you're interpreting as a Wheatstone Bridge configuration really isn't. There's an 'external' connection to three points on that bridge (node B is the culprit).

Instead, consider what would happen if you were to place a voltage source across AB; say, make A the positive end, and use B as the reference potential. What can you say about the potentials at the top and bottom rails relative to the reference node?

Staff: Mentor

I was thinking if the 10 ohm resistor (horizontal position in the image) should be removed, because if we analyse it from Wheatstone bridge's point of view, no current flows through it. I maybe wrong. Help appreciated.

You definitely cannot replace that resistor with an open circuit as you propose, that would alter the circuit.

You could, though, replace that resistor with a pair of appropriately-chosen resistors in parallel, and in that way effectively not alter the circuit at all.

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Staff: Mentor

It's commendable that, as an exercise, you'd choose to tackle this by the most complicated method. But why don't you also now solve it using one of the much simpler techniques suggested above so you can compare results?

For example, the very last computation involving the evaluation of a complicated fraction:

[tex]\frac{\frac{83}{6}\frac{56}{6}}{\frac{83+56}{6}}[/tex]

You got 33.4, but it actually evaluates to 5.57, neither of which is the right answer. This is simple carelessness.

But have a look at the last schematic to the left of that fraction. Where you have 56/6 in the fraction the schematic has 65/6; more carelessness. The complicated fraction should actually be:

[tex]\frac{\frac{83}{6}\frac{65}{6}}{\frac{83+65}{6}}[/tex]

which evaluates to 6.075. This result is better, but still not correct.

I also see a mistake in your very first delta-wye transformation. You have got 4 resistors which have become 10/3 ohms, but only 3 resistors should be affected by the transformation. The resistor which is at the bottom right of the original schematic should remain 10 ohms, not 10/3 ohms.

There may be more mistakes, but those are the ones I see at first glance.

The extreme symmetry of this circuit allows a simplification so that you don't need to do any delta-wye transformations; simple series and parallel combinations are enough.

Note that the very top node where 3 resistors are connected is at the same potential as the very bottom node where 3 other resistors are connected, so those two nodes can be connected together without any change in the A-B resistance.

You can do this by imagining a horizontal line right through the middle of the circuit. Fold the bottom half of the circuit along that line, up and over so that it then lies over the top half. Then the three 10 ohm resistors in the top half of the circuit become 5 ohm resistors, and the three resistors in the bottom half of the circuit vanish. Now you can easily evaluate the A-B resistance.